TITLE: Minimal "subset" of a set of homogeneous polynomials with same solution space
QUESTION [5 upvotes]: Suppose $A:=\{f_1,\dots,f_m\}\subset \mathbb{C}[x_1,\dots,x_n]$ with $m>n$ is a set of homogeneous polynomials of equal degree $d>0$. Suppose further that the variety they define consists of a single point (i.e. the zero vector).

Is it true that there exists a subset $g_1,\dots,g_n$ of the $\mathbb{C}$-linear span of $f_1,\dots,f_m$ with the same property? I.e. they have exactly one common zero?

That this is true for $d=1$ is basic linear algebra. For general $d$, my geometric intution (which, I admit, is probably rather bad given my inexperience in algebraic geometry), tells me that any subset $g_1,\dots,g_n$ "in general position" should have the desired property, which is why I believe this statement to be true. 
Thus, two "bonus questions":

Is it true that any "sufficiently generic" subset $g_1,\dots,g_n$ does it, and that "almost all" subsets of size $n$ are "sufficiently generic" in some suitable sense?

Finally (this holds for $d=1$, I think):

Can one always choose $g_1,\dots,g_n\in A$?

I wouldn't be surprised if for somebody who is familiar with algebraic geometry, this is a basic exercise, and so I apologize if this is deemed unsuitable for MO. I'd still be grateful for any references, proofs or hints.

REPLY [2 votes]: While subsets of $A$ will not work as Tony pointed out, the other questions have a positive answer. There are more general statements, but you situation is particularly vivid in the geometric sense (of course you need an infinite field, but I will assume algbraically closed for simplicity). Your condition implies that the set $A$  gives a morphism $\mathbb{P}^{n-1}\to\mathbb{P}^{m-1}$. If $m>n$, then the image is not all of $\mathbb{P}^{m-1}$ and so choose a point not in the image and project. We get a morphism $\mathbb{P}^{m-2}$. Instead of a point, you could choose a linear subspace of maximal dimension not intersecting the image and then you will end up with a map $\mathbb{P}^{n-1}$ to itself, giving you  what you need.